# Existence of iid random variables in Law of Large Numbers

Background info: I've taken a measure theory course, and I've read up on the basics of measure-theoretic probability theory. I'm trying to understand how to rigorously define iid random variables in a practical example.

Suppose we conduct an experiment to estimate the expected value of a random variable $X$ which measures the height of a population. To frame this all rigorously, let's say our probability space $(\Omega, \mathcal{F}, P)$ consists of a finite number of people, together with the uniform distribution, e.g. for $\omega_1, \omega_2 \in \Omega$, we have $P(\omega_1)=P(\omega_2)$. Define $X(\omega)$ to be the height of person $\omega$. In our experiment, at random we pick $n$ people from the population and measure their height, representing each individual person's height with the random variable $X_i$.

The Law of Large Numbers says if $X_i \sim X$ and the $X_i$ are iid, then the sample average tends to $E[X]$. My question, though, is how do I define the $X_i$? Each $X_i$ should represent a different person's height-- but if each $X_i$ is an indicator variable, then the $X_i$ are not the same distribution as $X$. It might make sense to set $X_i=X$, since then the $X_i$ are identically distributed, but then the $X_i$ are not independent.

I've seen posts that prove the existence of iid sequences in probability spaces, usually involving some infinite product of spaces, but I want an example of iid variables in a concrete setting. I'm currently studying some probability theory, and I find it easiest to understand the abstract definitions by relating them to intuitive examples.

• I did not understand your statement "if each $X_i$ is an indicator variable, then the $X_i$ are not the same distribution as $X$." This problem seems to have nothing to do with indicator variables. Of course, you will have trouble finding an infinite iid sequence from a finite population. What if you just imagine an infinite sequence of coin flips? Commented Aug 21, 2016 at 7:07
• The $X_i$ are not indicator random variables, they are samples drawn from the population. Commented Aug 21, 2016 at 7:09
• But how do you define a sample drawn from the population? To invoke the LLN, It has to be another random variable with the same distribution as $X$. But other random variables on our sample space have the same distribution as $X$? Commented Aug 21, 2016 at 7:12
• Michael: I guess for the LLN you want an infinite iid sequence, so sure, we could talk about coin flips instead. I know it doesn't have to do with indicator variables, but it was one of my ideas for defining the $X_i$. Commented Aug 21, 2016 at 7:15
• A framework to get the LLN is to consider $\Omega=(\mathbb R_+)^\mathbb N$ endowed with a product measure $P=\nu^\mathbb N$ and, for every $i$, $X_i(\omega)=\omega_i$. Then $(X_i)$ is i.i.d. with distribution $\nu$. Any finite (or finite countable) $\Omega$ fails this since the space of values of $(X_i)$ i.i.d. must be (uncountably) infinite (except if $P(X_1=x)=1$ for some $x$). Note finally that, as already explained elsewhere on the site, being preoccupied by the exact identity of $\Omega$ (as, unfortunately, many misguided curricula suggest) is a pure waste of time.
– Did
Commented Aug 21, 2016 at 9:39

Here's a concrete example of iid random variables. Let $X$ be a random variable uniformly distributed on the interval $(0,1)$. Let $X_i$ be the $i$th digit in the (non-terminating) binary expansion of $X$. Clearly $P[X_i=0]=P[X_i=1]=1/2$. Moreover, $X_1,X_2,X_3,\ldots$ are mututally independent.
• I'm struggling with understanding this. My situation is very similar. I consider $(0,1)$ with Lebesgue measure and the Borel sets. Then for $n\geq 1$ I let $X_n(x) := 1$ if the $n$'th digit in the binary expansion of $x$ is $1$ and $X_n(x) := -1$ if it is $0$. How do I show the $X_n$ are measurable? How do I show any finite number of them are independent? Commented Jan 15, 2023 at 20:36
• The set of points $x\in(0,1)$ whose $n$th binary digit is 1 is the union of $2^{n-1}$ intervals (each of length $2^{-n}$). This shows that the event $\{X_n=1\}$ is a Borel subset of $(0,1)$. Commented Jan 16, 2023 at 3:07